The correct stability order for N_2 and its g | vedclass

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(A) The stability of a molecule or ion is directly proportional to its bond order. For $N_2$ ($14$ electrons): Bond order = $3.0$. For $N_2^+$ ($13$ e

Vedclass · Accepted Answer

$N_2 > N_2^+ > N_2^- > N_2^{2-}$

Vedclass · Answer

(A) The stability of a molecule or ion is directly proportional to its bond order. For $N_2$ ($14$ electrons): Bond order = $3.0$. For $N_2^+$ ($13$ electrons): Bond order = $2.5$. For $N_2^-$ ($15$ electrons): Bond order = $2.5$. For $N_2^{2-}$ ($16$ electrons): Bond order = $2.0$. Since $N_2^+$ and $N_2^-$ have the same bond order,we compare the number of electrons in antibonding molecular orbitals $(ABMO)$. $N_2^+$ has $1$ electron in $ABMO$,while $N_2^-$ has $2$ electrons in $ABMO$. Stability is inversely proportional to the number of electrons in $ABMO$ when bond orders are equal. Therefore,the stability order is $N_2 > N_2^+ > N_2^- > N_2^{2-}$.

The correct stability order for $N_2$ and its given ions is

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